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lgbasallote Group Title

Prove that if n is a perfect square, then n +2 is not a perfect square

  • 2 years ago
  • 2 years ago

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  1. lgbasallote Group Title
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    i suppose first step would be to assume n is perfect square

    • 2 years ago
  2. lgbasallote Group Title
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    so n = m^2

    • 2 years ago
  3. lgbasallote Group Title
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    then i suppose i find n^2

    • 2 years ago
  4. lgbasallote Group Title
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    i mean n + 2

    • 2 years ago
  5. lgbasallote Group Title
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    n + 2 would be m^2 + 2

    • 2 years ago
  6. lgbasallote Group Title
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    so now... i assume m^2 + 2 is a perfect square...

    • 2 years ago
  7. mukushla Group Title
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    we have no 2 perfect square numbers with difference 2

    • 2 years ago
  8. lgbasallote Group Title
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    so i let m^2 + 2 = k^2

    • 2 years ago
  9. lgbasallote Group Title
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    @mukushla that's not really a proof.....

    • 2 years ago
  10. shubhamsrg Group Title
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    you may continue @igba

    • 2 years ago
  11. lgbasallote Group Title
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    hmm i suppose the next step would be m^2 - k^2 = 2 so (m+k)(m-k) = 2 i don't think this proves anything...

    • 2 years ago
  12. mukushla Group Title
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    actually that is :) \[m^2-k^2=2\]\[(m-k)(m+k)=2\]and this is impossible

    • 2 years ago
  13. lgbasallote Group Title
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    why so?

    • 2 years ago
  14. shubhamsrg Group Title
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    well,,you've almost proved it.. now m and k both are integers/.

    • 2 years ago
  15. shubhamsrg Group Title
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    and its k^2 -m^2 = 2 and not otherwise..

    • 2 years ago
  16. lgbasallote Group Title
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    oh yes

    • 2 years ago
  17. lgbasallote Group Title
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    2 = (k + m)(k-m)

    • 2 years ago
  18. lgbasallote Group Title
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    now...integers huh...

    • 2 years ago
  19. mukushla Group Title
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    \[m+k>m-k\]so\[m+k=2\]so\[m=k=1\]but it gives\[m-k=0\]

    • 2 years ago
  20. shubhamsrg Group Title
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    you know k>m.. so from (k+m)(k-m)= 1*2 =>k+m =2 and k-m = 1 this doesnt have an integral solution..

    • 2 years ago
  21. lgbasallote Group Title
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    why do i know k > m again?

    • 2 years ago
  22. shubhamsrg Group Title
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    m^2 + 2 = k^2 both m and k are positive integers hence k>m

    • 2 years ago
  23. shubhamsrg Group Title
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    in layman language,, k^2 - m^2 >0

    • 2 years ago
  24. lgbasallote Group Title
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    when was it assumed that m and k are positive?

    • 2 years ago
  25. shubhamsrg Group Title
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    what else do you mean by perfect square ?

    • 2 years ago
  26. lgbasallote Group Title
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    i don't see how that relates.... the definition of a perfect square is x = k^2

    • 2 years ago
  27. shubhamsrg Group Title
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    4 is a perfect sq since its 2^2 i.e. square of a positive integer.. and likewise..

    • 2 years ago
  28. estudier Group Title
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    @mukushla "we have no 2 perfect square numbers with difference 2" Because 0,1,4,9,6,5,6,9,4,1 etc....

    • 2 years ago
  29. lgbasallote Group Title
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    (-2)^2 is also 4...

    • 2 years ago
  30. shubhamsrg Group Title
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    you take the absolute value

    • 2 years ago
  31. shubhamsrg Group Title
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    i mean you have to..

    • 2 years ago
  32. lgbasallote Group Title
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    why?

    • 2 years ago
  33. shubhamsrg Group Title
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    see,, (k+m)(k-m) =2 either both are positive are both are negative,,since RHS is postive.. so even if you take -ve values,, negative * negative = positive ,,i.e. treated as absolute value..

    • 2 years ago
  34. lgbasallote Group Title
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    ...i don't think this is a valid proof....

    • 2 years ago
  35. shubhamsrg Group Title
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    another explanation might me x is a perfect square because its y^2 x = y^2 = (-y)^2 sqrt(x) = | y | = | -y |

    • 2 years ago
  36. jhonyy9 Group Title
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    who can writing here now two perfect squares with difference of 2 ? so because i think do not exist or .... ???

    • 2 years ago
  37. lgbasallote Group Title
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    since when did sqrt x become |-y|

    • 2 years ago
  38. estudier Group Title
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    0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats. No +2 in there.

    • 2 years ago
  39. jhonyy9 Group Title
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    so and i think that this difference of two will be the ,,key" of this proof of ...

    • 2 years ago
  40. jhonyy9 Group Title
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    @lgbasallote what is your opinion from this ,please ?

    • 2 years ago
  41. lgbasallote Group Title
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    my opinion is that this is some tricky algebra

    • 2 years ago
  42. jhonyy9 Group Title
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    how do you think it ???

    • 2 years ago
  43. jhonyy9 Group Title
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    so and why ?

    • 2 years ago
  44. lgbasallote Group Title
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    because i can't think of a proof

    • 2 years ago
  45. lgbasallote Group Title
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    how can you prove/disprove that the difference of two squares is 2...

    • 2 years ago
  46. lgbasallote Group Title
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    hmmm...wait...

    • 2 years ago
  47. jhonyy9 Group Title
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    so i think this is very easy logicaly but to prove it will be difficile i think so but there are again @saifoo.khan and @satellite73 probably they will can doing it hope so much

    • 2 years ago
  48. lgbasallote Group Title
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    their fields of specialty aren't exactly in proving

    • 2 years ago
  49. estudier Group Title
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    What's wrong with my proof?

    • 2 years ago
  50. lgbasallote Group Title
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    anyway... m^2 + 2 = k^2 k^2 - m^2 = 2 for it to be equal to 2...k^2 and m^2 should either be both even...or both odd so if i assume both are even.. then k and m are also both even so i can rewrite this as (2x)^2 - (2y^2) = 2 4x^2 - 4y^2 = 2 then... 2x^2 - 2y^2 = 1 x^2 - y^2 = 1/2 then since it's x^2 and y^2 are both integers...the difference should also be integer \('\therefore\) contradiction

    • 2 years ago
  51. lgbasallote Group Title
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    and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..

    • 2 years ago
  52. lgbasallote Group Title
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    @estudier i mean

    • 2 years ago
  53. lgbasallote Group Title
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    @mukushla too

    • 2 years ago
  54. estudier Group Title
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    So, that doesn't mean it's invalid..

    • 2 years ago
  55. lgbasallote Group Title
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    it does actually.....you can't do infinite substitutions to prove anything....

    • 2 years ago
  56. estudier Group Title
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    There is no infinity involved, all integer squares end in the numbers I gave.

    • 2 years ago
  57. lgbasallote Group Title
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    infinite loops involves infinity agree?

    • 2 years ago
  58. estudier Group Title
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    No infinite loop either, just logic.

    • 2 years ago
  59. estudier Group Title
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    An integer ends in one of digits 0 to 9. The squares must end in etc....

    • 2 years ago
  60. lgbasallote Group Title
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    but we both know that setting values is illegal in proving

    • 2 years ago
  61. estudier Group Title
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    I am not setting any values.

    • 2 years ago
  62. lgbasallote Group Title
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    and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever

    • 2 years ago
  63. lgbasallote Group Title
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    you're checking the squares of each number

    • 2 years ago
  64. estudier Group Title
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    I am not.

    • 2 years ago
  65. lgbasallote Group Title
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    and though there is a pattern...there is still susbtitution involved

    • 2 years ago
  66. lgbasallote Group Title
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    and then you'll have to substitute forever to verify that the pattern really does not break

    • 2 years ago
  67. lgbasallote Group Title
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    then infinity is involved

    • 2 years ago
  68. estudier Group Title
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    It's not a pattern, it's a fact 0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1

    • 2 years ago
  69. lgbasallote Group Title
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    so you do admit that you checked the 0 to 9 squares

    • 2 years ago
  70. estudier Group Title
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    Anything else is impossible.

    • 2 years ago
  71. lgbasallote Group Title
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    ...that's substitution

    • 2 years ago
  72. lgbasallote Group Title
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    anyway...my point is...substitution isn't really allowed in proving

    • 2 years ago
  73. estudier Group Title
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    You will find that this fact is used a lot in many valid proofs in number theory.

    • 2 years ago
  74. lgbasallote Group Title
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    perhaps

    • 2 years ago
  75. estudier Group Title
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    OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.

    • 2 years ago
  76. lgbasallote Group Title
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    since algebra is applicable in all numbers set to it....wouldn't that make it better than brute force?

    • 2 years ago
  77. estudier Group Title
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    I am not using brute force, brute force is something like testing or experimental math. Or exhaustion (which is also a valid proof technique). It's the use of a fact about squares (ALL squares)

    • 2 years ago
  78. lgbasallote Group Title
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    can you prove it? that it's a fact about squares?

    • 2 years ago
  79. estudier Group Title
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    Yes, all numbers end in 0 to 9 and so their squares end in .....QED.

    • 2 years ago
  80. lgbasallote Group Title
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    you didn't really demonstrate anything...you just repeated what you said

    • 2 years ago
  81. estudier Group Title
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    You asked for proof, I just gave u a proof.

    • 2 years ago
  82. estudier Group Title
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    If you deny my proof, provide a counterexample.

    • 2 years ago
  83. lgbasallote Group Title
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    since you try so hard to defend your proof, i'll accept it

    • 2 years ago
  84. estudier Group Title
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    That's very gracious, thank you:-)

    • 2 years ago
  85. lgbasallote Group Title
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    do check on my algebra proof though please. i would like to know if that's the way contradiction works

    • 2 years ago
  86. estudier Group Title
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    OK, I will look at it now....

    • 2 years ago
  87. estudier Group Title
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    Did u do the case when both are odd? (Perhaps a more straightforward way is one you start with n= some k squared then next up must be (k+1)^2 = k^2 + 2k +1 and see what you can deduce from that...)

    • 2 years ago
  88. lgbasallote Group Title
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    why do i have to do odd too? if i already disproved using even...doesn't that prove by contradiction already?

    • 2 years ago
  89. estudier Group Title
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    No because you divided the problem in cases, means you have to prove both (else you are missing all those cases)

    • 2 years ago
  90. lgbasallote Group Title
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    oh i see. but that kind of proof is right? because i'm a little skeptic if stating "2 can only be an outcome of even and even or odd and odd" is a valid statement

    • 2 years ago
  91. lgbasallote Group Title
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    again i ask... how is k > m

    • 2 years ago
  92. estudier Group Title
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    (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

    • 2 years ago
  93. mukushla Group Title
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    factoring involves lesser steps...its obvious that k>m \[k^2−m^2=2\]\[(k−m)(k+m)=2\]easily u can see \[k-m=1\]\[k+m=2\]add them\[2k=3\] Contradiction

    • 2 years ago
  94. lgbasallote Group Title
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    again i ask for the nth time @mukushla how is k > m

    • 2 years ago
  95. mukushla Group Title
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    ok man i just want to answer ur question

    • 2 years ago
  96. estudier Group Title
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    "2 can only be an outcome of even and even or odd and odd" is a valid statement Why do you think it might not be valid?

    • 2 years ago
  97. lgbasallote Group Title
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    well because 8 - 4 is not 2 but it's even minus even

    • 2 years ago
  98. lgbasallote Group Title
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    @mukushla i really don't get why k > m... i know k^2 > m^2 but does that make k > m?

    • 2 years ago
  99. estudier Group Title
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    OK, so finally, the algebra Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

    • 2 years ago
  100. mukushla Group Title
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    i thought they are positive :) sorry

    • 2 years ago
  101. lgbasallote Group Title
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    why (k+1)^2 @estudier ?

    • 2 years ago
  102. estudier Group Title
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    I already, said, that's the next square....

    • 2 years ago
  103. lgbasallote Group Title
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    next square?

    • 2 years ago
  104. lgbasallote Group Title
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    is this still related to my proof?

    • 2 years ago
  105. estudier Group Title
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    ? n is k^2, what is the next square?

    • 2 years ago
  106. mukushla Group Title
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    no problem ... this is better\[|k|^2-|m|^2=2\]\[(|k|-|m|)(|k|+|m|)=2\]so\[|k|-|m|=1\]\[|k|+|m|=2\]because this time im sure that \(|k|>|m|\) ...now add them\[2|k|=3\]Contradiction and we are done

    • 2 years ago
  107. estudier Group Title
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    No, this proof Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

    • 2 years ago
  108. lgbasallote Group Title
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    are you allowed to do that? @mukushla

    • 2 years ago
  109. lgbasallote Group Title
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    can you write it in a vertical fashion @estudier ? the signs are confusing me where the steps end

    • 2 years ago
  110. mukushla Group Title
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    why not...for any real number \[a^2=|a|^2\]

    • 2 years ago
  111. estudier Group Title
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    Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 k^2 + 2k +1 >= n +3 n +3 > n+2 QED

    • 2 years ago
  112. estudier Group Title
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    That's 2 proofs I have given now...

    • 2 years ago
  113. lgbasallote Group Title
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    a few questions in mind @estudier 1) why find the next square? 2) how is k^2 + 2k + 1 > = n + 3 3) what does n + 3 > n+2 prove?

    • 2 years ago
  114. estudier Group Title
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    Think about it....:-)

    • 2 years ago
  115. lgbasallote Group Title
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    in the meanwhile... @mukushla why |k| - |m| = 1?

    • 2 years ago
  116. mukushla Group Title
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    because\[|k|-|m|<|k|+|m|\]

    • 2 years ago
  117. lgbasallote Group Title
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    how does that make |k| - |m| = 1

    • 2 years ago
  118. mukushla Group Title
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    u have 2 positive integers with ab=2 and a<b what are a and b....clearly 1 and 2

    • 2 years ago
  119. mukushla Group Title
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    right?

    • 2 years ago
  120. lgbasallote Group Title
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    who said those are positive?

    • 2 years ago
  121. mukushla Group Title
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    |k|>|m| right?

    • 2 years ago
  122. lgbasallote Group Title
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    something feels wrong...

    • 2 years ago
  123. lgbasallote Group Title
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    anyway...where did you get 2|k| > 3

    • 2 years ago
  124. mukushla Group Title
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    ok...think on what i said...because i believe thats the shortest way to prove ur statement

    • 2 years ago
  125. lgbasallote Group Title
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    |2k| = 3 *

    • 2 years ago
  126. mukushla Group Title
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    adding\[|k|-|m|=1\]and\[|k|+|m|=2\]

    • 2 years ago
  127. estudier Group Title
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    Going back to your question about 2, u can just adjust it to difference of 2q...

    • 2 years ago
  128. lgbasallote Group Title
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    im really skeptical about this... especially that |k|^2 - |m|^2 = 2

    • 2 years ago
  129. lgbasallote Group Title
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    difference of 2q?

    • 2 years ago
  130. estudier Group Title
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    You were worried about the validity of your statement, change it to 2q instead of just 2.

    • 2 years ago
  131. estudier Group Title
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    Or 2p, or whatever letter u like best.

    • 2 years ago
  132. estudier Group Title
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    1) why find the next square? Because we want to see that it isn't n+2 2) how is k^2 + 2k + 1 > = n + 3 Substitute n=k^2 3) what does n + 3 > n+2 prove? See 1)

    • 2 years ago
  133. lgbasallote Group Title
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    oh. when i said difference of two numbers equal to 2

    • 2 years ago
  134. lgbasallote Group Title
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    but wait...if n + 2 were two squares away....wouldn't that make finding the next square invalid?

    • 2 years ago
  135. lgbasallote Group Title
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    and @mukushla how did you get |k| > |m|

    • 2 years ago
  136. estudier Group Title
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    No, because the next square is (k+1)^2 (which might or might not be equal to n+2, that is what we have to discover.

    • 2 years ago
  137. lgbasallote Group Title
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    and since the next square is greater than n + 2...that makes n + 2 invalid for any succeeding squares?

    • 2 years ago
  138. estudier Group Title
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    Yes, the algebra is completely general...

    • 2 years ago
  139. lgbasallote Group Title
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    i'm still confused about k^2 + 2k + 1 though...if n = k^2...it just makes n + 2k + 1...

    • 2 years ago
  140. estudier Group Title
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    ...which is greater than n+3, right?

    • 2 years ago
  141. lgbasallote Group Title
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    it is?

    • 2 years ago
  142. estudier Group Title
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    Number theorists don't pay attention to 0

    • 2 years ago
  143. lgbasallote Group Title
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    unless k is 0....

    • 2 years ago
  144. lgbasallote Group Title
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    why not?

    • 2 years ago
  145. estudier Group Title
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    Tradition, convention, call it as u like..... So in fact there are 2 definitions for natural number , one with 0 and one without.

    • 2 years ago
  146. sauravshakya Group Title
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    Can I try?

    • 2 years ago
  147. lgbasallote Group Title
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    ..you give so many weird proofs......the sad part is that i cannot disprove it because i know nothing about it....

    • 2 years ago
  148. estudier Group Title
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    More the merrier.

    • 2 years ago
  149. estudier Group Title
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    "because i know nothing about it" But you do now:-)

    • 2 years ago
  150. lgbasallote Group Title
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    it's not like it's limited to the 99s....

    • 2 years ago
  151. lgbasallote Group Title
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    i was referring to your claims earlier like brute force is valid proof in number theory and that number theorists don't consider 0...i cannot disprove your proofs because i do not know if these claims are true or not

    • 2 years ago
  152. estudier Group Title
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    Number theory is less onerous than real number stuff because it is integer only. But it is actually this difference that causes all the difficulty, students are not accustomed to thinking in a more freewheeling way.

    • 2 years ago
  153. sauravshakya Group Title
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    Let n be a perfect square where n=m^2 then next consecutive perfect square number would be (m+1)^2 Now, (m+1)^2-m^2=2m+1 Since, 2m+1>2 as m is a set of natural numbers. Thus, if n is a perfect square..... n+2 cannot be a perfect square as the difference between two perfect squares is always more than 2.

    • 2 years ago
  154. estudier Group Title
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    Yes, that's similar to the one I gave...

    • 2 years ago
  155. estudier Group Title
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    "number theorists don't consider 0" http://en.wikipedia.org/wiki/Natural_number

    • 2 years ago
  156. estudier Group Title
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    http://mathforum.org/library/drmath/view/63510.html Using 0,1,4, etc...

    • 2 years ago
  157. jhonyy9 Group Title
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    ok,so who can answering this question ? ,,the line of perfect squares begin with number ... " what is this number ?

    • 2 years ago
  158. jhonyy9 Group Title
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    so and for this your question @lgbasallote i think that for n=m^2 and n+2 for being a perfect square so than n+2=m^2 +2 suppose m^2 +2 =x^2 like a perfect square so than m^2 = x^2 -2 so m1,2 = +/- sqrt(x^2 -2)

    • 2 years ago
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